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Rate laws (2)

Rate laws (2). Lec 3 Week 4 . Non elementary Rate Laws. A large number of both homogeneous and heterogeneous reactions do not follow simple rate laws. Examples of Homogeneous Reactions CO +Cl 2 COCl 2

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Rate laws (2)

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  1. Rate laws (2) Lec 3 Week 4

  2. Non elementary Rate Laws • A large number of both homogeneous and heterogeneous reactions do not follow simple rate laws. Examples of Homogeneous Reactions CO +Cl2 COCl2 • This reaction is first order with respect to carbon monoxide, three-halves order with respect to chlorine. and five-halves order overall

  3. Non elementary Rate Laws Example of Heterogeneous Reactions • In many pas-solid catalyzed reactions. It historically has been the practice to write the rate law in terms of partial pressures rather than concentrations. C6H5CH3+ H2 C6H6 +CH4

  4. Rate laws for reversible reactions • we shall consider this gas-phase reaction to be elementary and reversible:

  5. Rate laws for reversible reactions

  6. Rate laws for reversible reactions

  7. Remember

  8. Rate laws for reversible reactions

  9. Rate laws for reversible reactions

  10. The reaction rate constant (k) • The reaction rate constant k or the specific reaction rate is not truly a constant. • It is merely independent of the concentrations of the species involved in the reaction. • It is almost always strongly dependent on temperature. • It depends on whether or not a catalyst is present, • In gas-phase reactions, it may be a function of total pressure. • In liquid systems it can also be a function of other parameters, such as ionic strength and choice of solvent. These other variables normally exhibit much less effect on the specific reaction rate than temperature does.

  11. Arrhenius equation • for the purposes of the material presented here. it will assumed that kA, depends only on temperature. • it was the great Swedish chemist Arrhenius who first suggested that the temperature dependence of the specific reaction rate, kA, could be correlated by the following equation: • where A = pre-exponential factor or frequency factor • E = activation energy. J/mol or cal/mol • R = gas constant = 8.3 14 J/mol.K = 1.987 cal/mol.K • T= absolute temperature, K

  12. The activation energy, E, is determined experimentally by carrying out the reaction at several different temperatures. After taking the natural logarithm of Equation (Arrhenius equation) we obtain; • and see that the activation energy can be found from a plot of In kA, as a function of (1/T)

  13. Assignment • Example (1) Calculate the activation energy for the decomposition of benzene diazoninum chloride to give chlorobenzene and nitrogen: • using the information in Table (1) for this first-order reaction.

  14. Solution Graphically Semi log plot of 1/T vs. log K and use the following equation to determine (E) Rearrange to get

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